leetcode33

这道题目看了题目解答才做出来

旋转元素各不相同数组二分查找

自己代码：

public int search(int[] nums, int target)
{
    if(nums.length==0)return -1;
    if(nums.length==1&&nums[0]==target)return 0;
    if(nums.length==1&&nums[0]!=target)return -1;
    int low=0;
    int high=nums.length-1;
    int mid;
    while(true)
    {
        mid=low+((high-low)>>2);
        if(nums[mid]==target)return mid;
        else
        {
            if(nums[mid]>nums[low])
            {
                if(target>=nums[low]&&target<nums[mid])
                {
                    high=mid-1;
                    break;
                }
                else low=mid+1;
            }
            else
            {
                if(target>nums[mid]&&target<=nums[high])
                {
                    low=mid+1;
                    break;
                }
                else high=mid-1;
            }
        }
    }
    if(low>high)return -1;
    return Arrays.binarySearch(nums,low,high+1,target);
}

题目解答代码：

class Solution {
    public int search(int[] nums, int target) {
        int len = nums.length;
        int left = 0, right = len-1;
        while(left <= right){
            int mid = (left + right) / 2;
            if(nums[mid] == target)
                return mid;
            else if(nums[mid] < nums[right]){
                if(nums[mid] < target && target <= nums[right])
                    left = mid+1;
                else
                    right = mid-1;
            }
            else{
                if(nums[left] <= target && target < nums[mid])
                    right = mid-1;
                else
                    left = mid+1;
            }
        }
        return -1;
    }
}

反思：

• 为什么要用True循环呢？查找mid的过程本来就是和二分查找是类似的，所以直接low<=high就可以了。
• 不要想太多，不要把过程拆解来看，寻找有序的left和right，找完了之后还是要在这里面找target的呀！